Physics Formulas

Escape Velocity Formula

You must have seen rockets leaving the earth to go into space. Then you must have also noticed they require a very huge kick-start in order to leave the surface of the earth. It is because of the strong gravitational field of the surface of the earth. Thus, this is where escape velocity comes in. In this article, you will learn about the escape velocity formula and see how the formula can be used to find out the escape velocity of any object or body.

escape velocity formula


Escape Velocity is referred to as the minimum velocity needed by anybody or object to be projected to overcome the gravitational pull of the planet earth. In other words, the minimum velocity that one requires to escape the gravitational field is escape velocity.

Basically, it means escaping the land without any chance of falling back. Therefore, any object or body having escape velocity on the surface of the earth can totally escape the gravitational field of the earth in addition to avoiding the losses due to the atmosphere.

For instance, if you see that a spacecraft leaves the earth’s surface it requires a speed of 7 miles per second or about 25,000 miles per hour to leave the earth without ever falling back to the surface of the earth.

You can download Escape Velocity Cheat Sheet by clicking on the download button below

escape velocity formula

Escape Velocity Formula

Escape velocity refers to the minimum velocity which is needed to leave a planet or moon. For instance, for any rocket or some other object to leave a planet, it has to overcome the pull of gravity.

The formula for escape velocity comprises of a constant, G, which we refer to as the universal gravitational constant. The value of it is = 6.673 × 10-11 N . m2 / kg2. The unit for escape velocity is meters per second (m/s).

Escape velocity = \(\sqrt{\frac{2 (gravitational constant) (mass of the planet of moon) }{radius of the planet or moon}}\)

v escape = \(\sqrt{\frac{2GM}{R}}\)

Over here:
v escape refers to the escape velocity (m/s)
G is the universal gravitational constant (6.673 × 10-11 N . m2 / kg2)
M refers to the mass of the planet or moon (kg)
R is the radius of the planet or moon (m)

 Solved Question

Question– Suppose the radius of Earth is 6.38 × 106 m and the mass of the planet earth is 5.98 × 1024 kg. Find out the escape velocity from planet earth.

Answer– We can find the escape velocity from earth using the escape velocity formula:

vescape = \(\sqrt{\frac{2GM}{R}}\)
Thus, when we replace it with figures, we will get:
vescape = \(\sqrt{\frac{2(6.673 \times 10^{-11}N^{.}m^{2}/kg^{2})(5.98 \times 10^{24}kg)}{6.38 \times 10^{6}m}}\)
\(v_{escape} = \sqrt{\frac{7.981 \times 10^{14}N \cdot m^{2}/kg}{6.38 \times 10^{6}m}}\)
vescape = \(\sqrt{1.251\times 10^{8} N \cdot m^{2}/kg}\)
vescape = 11184 m/s

Therefore, the escape velocity from Earth is 11 184 m/s, or approximately 11.2 km/s.

Question– In order to leave the moon, the Apollo astronauts had to take off in the lunar mobile and reach the escape velocity of the moon. The radius of the moon is 1.74 × 106 m, and the mass of the moon is 7.35 × 1022 kg. Calculate the velocity which the Apollo astronauts have to reach in order to leave the moon?

Answer– We can find the escape velocity from the moon using the escape velocity formula:
vescape = \(\sqrt{\frac{2GM}{R}}\)
\(v_{escape} = \sqrt{\frac{2(6.673 \times 10 ^{-11}N \cdot m^{2}/kg^{2}) (7.35 \times 10^{22}kg)}{1.74 \times 10^{6}m}}\)
\(v_{escape} = \sqrt{\frac{9.809 \times 10 ^{-11}N \cdot m^{2}/kg}{1.74 \times 10^{6}m}}\)
\(v_{escape} = \sqrt{5.638 \times 10^{6}N \cdot m^{2}/kg}\)
$$v_{escape} = 2374$$ m/s.

Therefore, the escape velocity from the moon is 2374 m/s, or almost 2.37 km/s.

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Yashdeep tiwariKwame DavidumerMalek safrinRoger Carmichael Recent comment authors
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Roger Carmichael

Typo Error>
Speed of Light, C = 299,792,458 m/s in vacuum
So U s/b C = 3 x 10^8 m/s
Not that C = 3 x 108 m/s
to imply C = 324 m/s
A bullet is faster than 324m/s

Malek safrin

I have realy intrested to to this topic


m=f/a correct this

B. Akshaya
B. Akshaya


Kwame David

Interesting studies

Yashdeep tiwari
Yashdeep tiwari

It is already correct f= ma by second newton formula…

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